# lesson24

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```					A question about polygons

Devising a computational method
for determining if a point lies in a
plane convex polygon
Problem background
• Recall our „globe.cpp‟ ray-tracing demo
• It depicted a scene showing two objects
• One of the objects was a square tabletop
• Its edges ran parallel to x- and z-axes
• That fact made it easy to determine if the
spot where a ray hit the tabletop fell inside
or outside of the square‟s boundary edges
• We were lucky!
Alternative geometry?
• What if we wanted to depict our globe as
resting on a tabletop that wasn‟t a nicely
aligned square? For example, could we
show a tabletop that was a hexagon?
• The only change is that the edges of this
six-sided tabletop can‟t all be lined up with
the coordinate system axes
• We need a new approach to determining if
a ray hits a spot that‟s on our tabletop
A simpler problem
• How can we tell if a point lies in a triangle?
c

p                        q

b
a
Point p lies inside triangle Δabc       Point q lies outside triangle Δabc
Triangle algorithm
• Draw vectors from a to b and from a to c
• We can regard these vectors as the axes
for a “skewed” coordinate system
• Then every point in the triangle‟s plane
would have a unique pair of coordinates
• We can compute those coordinates using
Cramer‟s Rule (from linear algebra)
The algorithm idea
c
ap = c1*ab + c2*ac                      p

a                            b

c1 = det( ap, ac )/det( ab, ac )
c2 = det( ab, ap )/det( ab, ac )
Cartesian Analogy
y-axis

p = (c1,c2)

x-axis

x+y=1

p lies outside the triangle if c1 < 0 or c2 < 0 or c1+c2 > 1
Determinant function: 2x2
typedef float scalar_t;
typedef struct { scalar_t x, y; } vector_t;

scalar_t det( vector_t a, vector_t b )
{
return     a.x * b.y – b.x * a.y;
}
Constructing a regular hexagon
theta = 2*PI / 6

( cos(2*theta), sin(2*theta) )            ( cos(1*theta), sin(1*theta) )

( cos(3*theta), sin(3*theta) )                         ( cos(0*theta), sin(0*theta) )

( cos(4*theta), sin(4*theta) )             ( cos(5*theta), sin(5*theta) )
Subdividing the hexagon

A point p lies outside the hexagon -- unless
it lies inside one of these four sub-triangles
Same approach for n-gons
• Demo program „hexagon.cpp‟ illustrates
the use of our just-described algorithm
• Every convex polygon can be subdivided
into triangles, so the same ideas can be
applied to any regular n-sided polygon
• Exercise: modify the demo-program so it
draws an octagon, a pentagon, a septagon
Extension to a tetrahedron
• A tetrahedron is a 3D analog of a triangle
• It has 4 vertices, located in space (but not
all vertices can lie in the same plane)
• Each face of a tetrahedron is a triangle
c
b

o
a
Cramer‟s Rule in 3D
typedef float scalar_t;
typedef struct { scalar_t x, y, z; } vector_t;
scalar_t det( vector_t a, vector_t b, vector_t c )
{
scalar_t     sum = 0.0;
sum += a.x * b.y * c.z – a.x * b.z * c.y;
sum += a.y * b.z * c.x – a.y * b.x * c.z;
sum += a.z * b.x * c.y – a.z * b.y * c.x;
return       sum;
}
Is point in tetrahedron?
• Let o, a, b, c be vertices of a tetrahedron
• Form the three vectors oa, ob, oc and
regard them as the coordinate axes in a
“skewed” 3D coordinate system
• Then any point p in space has a unique
triple of coordinates:
op = c1*oa + c2*ob + c3*oc
• These three coordinates can be computed
using Cramer‟s Rule
Details of Cramer Rule
c1 = det( op, ob, oc )/det( oa, ob, oc )
c2 = det( oa, op, oc )/det( oa, ob, oc )
c3 = det( oa, ob, op )/det( oa, ob, oc )

Point p lies inside the tetrahedron – unless
c1 < 0 or c2 < 0 or c3 < 0
or             c 1 + c2 + c3 > 1
Convex polyhedron
• Just as a convex polygon can be divided
into subtriangles, any convex polyhedron
can be divided into several tetrahedrons
• We can tell if a point lies in the polyhedron
by testing to see it lies in one of the solid‟s
tetrahedral parts
• An example: the regular dodecahedron
can be raytraced by using these ideas!

```
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 views: 10 posted: 11/11/2011 language: English pages: 16